Given the very low value of K_sp Cr (OH)_3, a precipitate of Cr(OH)_3 would be e

Question

Given the very low value of K_sp Cr (OH)_3, a precipitate of Cr(OH)_3 would be expected if only 6 M NaOH were added to the mixture of Group III cations in the first step of the procedure. Explain how and why the chromium remains in solution. A solution may contain one or more of the Group III cations. When this solution is combined with N_aOH (aq), N_aOCI(aq) and NH_3, (aq) only a colorless solution is obtained with no precipitate evident. Indicate whether each of the following cations is present, absent or undetermined. Cr^3+ AI^3+ Fe^3+ Ni^2+

Explanation / Answer

Q1.

Ksp low for Cr(OH)3

Note that initially:

Cr+3 + 3OH- --> Cr(OH)3

addition of extra OH- ions will form:

[Cr(H2O)6]3 --> Cr(H2O)3(OH)] --> [Cr(OH)6]3-

which are ALL aquous solution

Q2.

Cr+3 not likely, since it should have precipitate

Al+3 not likely, since it should have precipitate, Al(OH)3

Fe+3 not likely, since it should have precipitate, also ammonia should have changed colors

Ni2+ not likely, since it should have precipitate Ni(OH)2

Related Questions

Navigate

• Browse (All)
• Browse G
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.